The escape velocity from the Earth's surface is \(v\). The escape velocity from the surface of another planet having a radius, four times that of Earth and the same mass density is: 

A particle of mass \(m\) is projected with a velocity, \(v=kV_{e} ~(k<1)\) from the surface of the earth. The maximum height, above the surface, reached by the particle is: (Where \(V_e=\) escape velocity, \(R=\) radius of the earth)

A black hole is an object whose gravitational field is so strong that even light cannot escape from it. To what approximate radius would Earth (mass \(= 5.98\times 10^{24}~\text{kg}\)) have to be compressed to be a black hole?
1. \(10^{-9}~\text{m}\)
2. \(10^{-6}~\text{m}\)
3. \(10^{-2}~\text{m}\)
4. \(100​~\text{m}\)

A particle of mass \(\mathrm{m}\) is thrown upwards from the surface of the earth, with a velocity \(\mathrm{u}\). The mass and the radius of the earth are, respectively, \(\mathrm{M}\) and \(\mathrm{R}\). \(\mathrm{G}\) is the gravitational constant and \(\mathrm{g}\) is the acceleration due to gravity on the surface of the earth. The minimum value of \(\mathrm{u}\) so that the particle does not return back to earth is:
1. \(\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}^2}} \)
2. \(\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} \)
3.\(\sqrt{\frac{2 \mathrm{gM}}{\mathrm{R}^2}} \)
4. \(\sqrt{ \mathrm{2gR^2}}\)

The earth is assumed to be a sphere of radius R. A platform is arranged at a height R from the surface of the earth. The escape velocity of a body from this platform is fv_e, where v_e is its escape velocity from the surface of the earth. The value of f is:

1. $\sqrt{2}$

2. $\frac{1}{\sqrt{2}}$

3. $\frac{1}{3}$

4. $\frac{1}{2}$

For a planet having mass equal to the mass of the earth but radius equal to one-fourth of the radius of the earth, its escape velocity will be: